网上一个经典异步FIFO程序

xuduo6845

请问其中ASIZE为什么是4?我觉得应该是3.我在一本书上看到也是4.
/ D+ v% j4 i& N, Gmodule fifo1(rdata, wfull, rempty, wdata, winc, wclk, wrst_n,rinc, rclk, rrst_n);
$ r+ d( L4 t1 e9 k, Z" B& Pparameter DSIZE = 8; parameter ASIZE = 4;
! \- s  x% v$ Foutput [DSIZE-1:0] rdata;
output wfull;
6 n6 ^; J7 b: u7 u! Loutput rempty;
  T2 q. s( |5 T  m: i* P5 b' n1 kinput [DSIZE-1:0] wdata;
% h" y) S) m# t, W7 B  B; xinput winc, wclk, wrst_n;
( c0 C3 P" K* \% v/ sinput rinc, rclk, rrst_n;
7 u( j8 i, B7 G- Greg wfull,rempty;
+ _9 p7 C. I# K9 p% T8 g  F. Oreg [ASIZE:0] wptr, rptr, wq2_rptr, rq2_wptr, wq1_rptr,rq1_wptr;
9 A8 }1 P2 D5 X0 xreg [ASIZE:0] rbin, wbin;
9 f/ }5 R3 y1 U, B2 }, u( [reg [DSIZE-1:0] mem[0: (1<<ASIZE)-1];
wire [ASIZE-1:0] waddr, raddr;
0 R2 l9 k& q. \* o1 K, @8 hwire [ASIZE:0] rgraynext, rbinnext,wgraynext,wbinnext;
wire rempty_val,wfull_val;
0 f1 Z0 @$ k+ j3 B1 g- |' H
( f. }" Q7 w1 d7 W1 a1 ]. [assign rdata=mem[raddr];
always@(posedge wclk)
if (winc && !wfull) mem[waddr] <= wdata;
- O+ E& V0 O9 |$ m& I
3 g, ?# g1 I+ P) X; z3 oalways @(posedge wclk or negedge wrst_n)
if (!wrst_n) {wq2_rptr,wq1_rptr} <= 0;
else {wq2_rptr,wq1_rptr} <= {wq1_rptr,rptr};

always @(posedge rclk or negedge rrst_n)
1 Z) Q2 Y/ c- m$ g. c3 Xif (!rrst_n) {rq2_wptr,rq1_wptr} <= 0;
else {rq2_wptr,rq1_wptr} <= {rq1_wptr,wptr};

3 c$ ?5 ]: J( _- ?1 K
always @(posedge rclk or negedge rrst_n) // GRAYSTYLE2 pointer
" M+ j% {! s$ ^5 t4 P: Ibegin
. {& H9 f) C: e: \; p; Pif (!rrst_n) {rbin, rptr} <= 0;
else {rbin, rptr} <= {rbinnext, rgraynext};
end
0 b: O/ f3 i/ j& L4 ]4 D

8 G6 [0 W% Z8 P- Q" b0 o0 jassign raddr = rbin[ASIZE-1:0];
assign rbinnext = rbin + (rinc & ~rempty);
5 U) }" U- [. Z. ~$ i* X* v- Fassign rgraynext = (rbinnext>>1) ^ rbinnext;
" S1 g/ f: t9 ?; C, X
$ n9 O$ P0 Q2 G8 f% p
; ]" v, s; B6 M$ a# Vassign rempty_val = (rgraynext == rq2_wptr);
) j" ~3 b9 ], V/ S6 E* salways @(posedge rclk or negedge rrst_n)
begin
* C- t/ u8 I& D8 hif (!rrst_n) rempty <= 1'b1;
else rempty <= rempty_val;
4 q+ ?4 x& q! k! \end


always @(posedge wclk or negedge wrst_n) // GRAYSTYLE2 pointer
" C" S8 i3 j* o- S4 A) ~& {if (!wrst_n) {wbin, wptr} <= 0;
else {wbin, wptr} <= {wbinnext, wgraynext};

6 a7 d  c( q1 n& g/ `$ M
assign waddr = wbin[ASIZE-1:0];
assign wbinnext = wbin + (winc & ~wfull);
assign wgraynext = (wbinnext>>1) ^ wbinnext;
assign wfull_val = (wgraynext=={~wq2_rptr[ASIZE:ASIZE-1], wq2_rptr[ASIZE-2:0]}); //:ASIZE-1]
always @(posedge wclk or negedge wrst_n)
if (!wrst_n) wfull <= 1'b0;
8 ?3 L4 d$ u9 [else wfull <= wfull_val;
! `" h/ N. C- }3 t8 ]endmodule

, o* f4 u: {) {, L' u( ^[ 本帖最後由 xuduo6845 於 2009-9-4 11:25 AM 編輯 ]

addn

您好

9 c+ I- U1 I, N" d這個好久前有大概看過,現在忘了差不多了,只剩模糊印像,
0 `0 \  F4 R. a! g/ WASIZE = 4,實際memory address為ASIZE-1
- o$ j+ \8 E/ a/ f  \主要是這做法用了反射式gray 碼的原故,所以memory address為4bit
9 u, k0 W: r  n, t轉換為反射式gray 碼就要變成5bit,最高bit可被用來做判斷.......細節你
0 q1 x; T! P# P3 B/ t0 m" @7 C要在找看看,抱歉我只記得這樣

tommy123

多一個bit主要是要用來計算 FIFO Full or Empty